Match the LIST-I with LIST-II.
| LIST-I Reaction | LIST-II Most appropriate reaction order |
| A. \( \ce{2NH3(g)} \xrightarrow[\text{Pt catalyst}]{\text{1130K}} \ce{N2(g) + 3H2(g)} \) | I. First-order Reaction |
| B. \( {}^{226}_{88}\mathrm{Ra}\rightarrow{}^{4}_{2}\mathrm{He}+{}^{222}_{86}\mathrm{Rn} \) | II. Second-order Reaction |
| C. KClO3 + 6FeSO4 + 3H2SO4 → KCl + 3Fe2(SO4)3 + 3H2O | III. Zero-order Reaction |
| D. \( \ce{C12H22O11 + H2O} \xrightarrow{\ce{H+}} \ce{C6O12O6 + C6O12O6} \) | IV. Pseudo first-order Reaction |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → A-III, B-I, C-II, D-IV
The order of a reaction is the sum of the powers of the concentration terms in the experimental rate law. It is found by experiment and cannot be read from the balanced equation. Four typical cases are:
Zero order: rate does not depend on concentration, e.g. a gas decomposing on a metal surface that is fully covered.
First order: rate ∝ one concentration term, e.g. all radioactive decays.
Second order: rate ∝ two concentration terms, even when the equation shows many molecules, if the reaction occurs in steps.
Pseudo first order: a higher order reaction in which one reactant is in large excess.
A. \( \ce{2NH3(g)} \xrightarrow[\text{Pt catalyst}]{\text{1130K}} \ce{N2(g) + 3H2(g)} \) — This matches III. Zero-order Reaction. At high pressure the platinum surface is saturated with \( \ce{NH3} \), so a change in the gas concentration cannot change the rate.
\( \displaystyle \text{Rate}=k[\ce{NH3}]^0=k \)
B. \( {}^{226}_{88}\mathrm{Ra}\rightarrow{}^{4}_{2}\mathrm{He}+{}^{222}_{86}\mathrm{Rn} \) — This matches I. First-order Reaction. Every nuclear decay is first order: a fixed fraction of the nuclei present decays in a given time.
\( \displaystyle \text{Rate}=k[\ce{Ra}] \)
C. \( \ce{KClO3} \) + \( \ce{6FeSO4} \) + \( \ce{3H2SO4} \) → KCl + \( \ce{3Fe2(SO4)3} \) + \( \ce{3H2O} \) — This matches II. Second-order Reaction. Ten molecules appear on the left, so it seems to be of tenth order, but it is a complex reaction that takes place in several steps; experiments show it is second order.
D. \( \ce{C12H22O11 + H2O} \xrightarrow{\ce{H+}} \ce{C6O12O6 + C6O12O6} \) — This matches IV. Pseudo first-order Reaction. In the inversion of cane sugar into glucose and fructose, water is in large excess, so its concentration stays constant and only \( \ce{[C12H22O11]} \) appears in the rate law.
\( \displaystyle \text{Rate}=k[\ce{C12H22O11}] \)
Shortcut: The pair A–III appears only in Option (1).
Note: It seems that there is a mistake in the original NTA paper, which prints the products in item (D) as C6O12O6 + C6O12O6 when it should read C6H12O6 + C6H12O6 (glucose + fructose).